Комплекс сонлар

2.4. Kompleks sonlarga oid ayrim tipik misollar
1) x,yR deb faraz qilib, x+8i+(y-3)i=1 tenglamadan x va y larni toping.
x+8i+yi-3i=1x+yi=l-5i  x=l, y=-5.
2) ni hisoblang.

3) z₁ = a₁ +b₁i, z₂=a₂+b₁i, bo‘lsa bo‘lishini isbotlang.

bulardan kelib chiqadi.
4) amallarni bajaring.

5) z²+ =0 tenglamani eching.
bo‘lsin, u holda =a-bi bo‘ladi.
(a+bi)²+(a-bi)=0 a²+2abi-b²+a-bi=0 
(a²+a-b²)+(2ab-b)i=0 

1. 
   b=0 bo‘lganda, a²+a=0(a₁=0  a₂=-1)
   
2. 
   bo‘lganda,
   

Demak, z₁=0, z₂=-1, z₃= z₄= , lar berilgan tenglamaning ildizlari bo‘ladi.
6) z= hisoblansin.
sign (-10)=-l bo‘lgani uchun quyidagiga ega bo‘lamiz:

7) z=1-i sonini trigonometrik shaklda yozing.
a=1, b=-1 bo‘lgani uchun

8) Agar
bo‘lsa, ni hisoblang.

ni hisoblash uchun z₁ ni trigonometrik shaklga keltiramiz.
(SHu kabi ).
chunki, , shu kabi . Demak

9) hisoblansin.
 a=-2, b=2 bo‘lgani sababli r= , cos = . Bundan

10) 1-C₈²+ C₈⁴- C₈⁶+ C₈⁸=16 ayniyatni hisoblang.
A (l+i)⁸=l+i C₈¹ - C₈² - iC₈³+ C₈⁴+ iC₈⁵ - C₈⁶ - iC₈⁷+ C₈²+ C₈⁸=
=(1- iC₈² + C₈⁴ - C₈⁶+ C₈⁸)+ i(C₈¹ - C₈³ +C₈⁵ -C₈⁷) (*)
(1+i)⁸=₁=
Kompleks sonlarning tenglik shartini e’tiborga olib, (*) va (**) lardan 1- i C₈² + C₈⁴ - C₈⁶+ C₈⁸=16, C₈¹ - C₈³ +C₈⁵+ C₈⁷=0 kelib chiqadi.
11) cos3 va sin3 larni cos va sin larning darajalari orqali ifodalang.
 (cos+sin)³= cos3+isin3 ikkinchi tomondan (cos+sin)³)³= (cos+sin)³= cos³j+3isinjcos²j-3sin²jcosj-isin³j/
Bulardan cos3j=cos³j-3sin²jcosj, sin3j=3sinjcos²j-sin³j.
12) ni hisоblаng.
Yechish. Аvvаl ni trigоnоmеtrik shаklgа kеltirib оlаmiz:
.
Hоsil bo‘lgаn trigоnоmеtrik shаkldаgi kоmplеks sоndаn 3-dаrаjаli ildizlаrni fоrmulа yordаmidа tоpаmiz:

Хаr bir ildizni аlоhidа ifоdаlаymiz:

13) (5-3i) + (3+3i)=(5+3) + (3-3)i= 8
(2+5i) + (-2+5i)=(2-2) + (5+5)i= 10i
14) (10+2i) – (3-4i)= (10-3) – (2+4)i= 7+6i
(4+5i) – (3+5i)= (4-3) – (5-5)i= 1
15) (5+2i)(3-4i)= 23-14i
(2+i)(2-i)= 4+1=5
16)
17)Algebraik ko‘rinishdagi kompleks sonni trigonometrik ko‘rinishga o‘tkazish. α=1+i r=|1+i|= , , , demak, ;
α=1+i=


18) 2(Cos20⁰ + iSin20⁰) · 7(Cos100⁰ + iSin100⁰)=
= 14(Cos120⁰ + iSin120⁰)=